Welcome back! The following blog pages are prerequisites for understanding this page:
This page is challenging, not only to write, but also to fully comprehend. Feel free to
pepper me with questions about any of the steps shown below. I will attempt to answer all such
questions.
If you missed the first page of this puzzle proof, it can be found
here.
Standard Depth 4 forbidding chain using candidates 135
Illustrated to the left:
- d3=1 == d1=1 -- d1=3 == d5=3 -- c5=3 == c2=3 -- c2=5 == c3=5
This is nothing but a typical forbidding chain. It is heavily indicated by my standard puzzle
mark-up. Please remember that strong and weak inferences remain
not mutually exclusive.
Forbidding Chain using candidates 5 & 6
Above, another typical depth 4 forbidding chain is illustrated. Again, this one is not hard
to locate if one understands the concepts behind my standard puzzle markings.
- d7=6 == d6=6 -- d6=5 == d2=5 -- c2=5 == c3=5 -- c3=6 == g3=6
After this elimination, very few simple chains remain for one to find. Thus, the following
chains are best found after one has practiced quite a bit at finding
forbidding chains.
A forbidding chain that uses Almost Coloring and an Almost Locked Set
Oftentimes, eliminations such as the one above appear more clearly in my head than I can
illustrate them. The chain above uses an Almost Finned X wing on 4s at d67, i678 and
an Almost Naked Pair 34 at fi7, plus a couple of simple strong inferences to connect the dots.
- {pair 34 at fi7} == f7=2 -- f4=2 == e4=2 -- e4=3 == d5=3 -- d5=4_{1}
== {fc on 4s: d7 ==_{1} d6 -- i6 == i78
The subscripts attempt to identify the members of the strong set: d567=4. All three members of
that strong set must be included in the chain if one intends to use the strong inference between
any of them. This chain has a rather interesting and complex conclusion. Properly understood,
the chain concludes that:
- {d7=4 == i78=4} == {pair 34 at fi7}
- pair 34 at fi7 reduces to f7=4 == i7=4
- Since == means OR (also called union), one can get rid of the brackets:
- {d7=4, i7=4, f7=4, i8=4} is a strong set - at least one of them is true.
- Any candidate 4 seen by all 4 of the cells: dfi7, i8 can be eliminated.
The keys to finding such an elimination are:
- Quickly being able to indentify potential coloring eliminations such as a finned X wing
- Being able to unilyze a bridge between two techniques. Here the bridge is built
about a Hub at e4, with a strong link in 2s and 3s emanating from it.
- Understanding the power of Almost Locked Sets used in a chain.
- Understanding how the conclusions of the chain actually work.
This one step is plenty to study. But, the elimination does not do much. Yet. More chains
like this last one are required, IMHO.
Almost Coloring used in a Chain - Again....
Above, but for the 4 at g9, one would have a coloring elimination available of 4 from c8, using
the 4s in column g and the 4s in box b5. However, we cannot ignore that 4 at g9. Instead, we use
it:
- g9=4_{1} == {fc on 4s: g8 ==_{1} g4 -- a4 == c5} -- c8=4 == c8=6 -- c3=6 == g3=6
Note that the bracketed chain (fc on 4s...) reduces to g8 == c5. Since both c5, g8 see c8, the chain, if
it exists (meaning is true) forbids c8=4. However, if the chain does not exist, then the entry
condition for the chain, g9=4, must be true. In any event, one can safely determine that g9 cannot
be 6.
Some may claim that this chain is not bidirectional. This is not true. The chain makes perfect
sense the other way around:
- g3=6 == c3=6 -- c8=6 == c8=4 --{fc on 4's: c5 == a4 -- g4 ==_{1} g8} == g9=4_{1}
It may take some thinking to realize that c8=4 forbids the entire boolean represented by the
fc on 4's from existing. However, that is precisely the case. Then, it is clear that if that
Boolean is invalid, then clearly g9=4. It may take some study to fully understand this concept.
Do not be concerned. It can get harder!
Relatively easy use of an Almost Locked Set (pair 56) in a chain
Above, the relatively straightforward chain using an Almost Locked Set:
- e9=2 == {pair 56 at e69} -- e3=5 == d2=5 -- d2=2 == h2=2 => h9≠2
This one has been available for some time. I thought that I would insert it here between the harder
ones to give everyone a break.... Perhaps I have a twisted sense of humour?
An Almost AIC used within an AIC
The step illustrated above is not actually required to make the following step. However, it
significantly simplifies the logic. I am not sure if I could find the following step
without finding and using this one first. This Forbidding Chain can be written as:
- h9=6_{1} == {d6=6 == d7=6 -- e9=6 == _{1} a9=6 -- c8=6 == c8=4 -- c5=4 == d5=4} -- d6=4 == i6=4 -- i6=5 == i8=5
Here, both ends of the bracketed chain forbid d6=4. Thus the chain reduces to:
- h9=6 == MESS -- d6=4 == i6=4 -- i6=5 == i8=5
This can be further reduced to simply: h9=6 == i8=5.
This chain is not really any different than using Almost Locked Sets, or Almost Coloring, within
a forbidding chain, or Alternating Inference Chain. The only problem is finding it. That
challenge may be better treated in a future blog page. For now, trust that once one gets very
good at finding smaller less complex chains, the bigger ones with complex interactions become
less intimidating.
Locked candidate 5
Needing a little break from that last one, either one of two ways can be used to get f8≠5
This concludes the second page of this puzzle proof. More possible headaches, can be found on the
next page.